Question

The two ends $A$ and $B$ of a uniform rod of length $l=1\text{m}$ are moving with velocities $V_A$ and $V_B$ as shown. The velocity $V_B\left(\text{m/s}\right)$ is

Answer 1

$15.00$
Given that,
Velocity of $A$ = $20\text{m/s}$
Velocity of $B$ = $V_B\text{m/s}$
According to diagram
Velocity of $A$ along $X-axis$ along $AB$ is
$V_{AX}=V_A\cos {53}^{\circ }$
$V_{AX}=20\cos {53}^{\circ }$
$V_{AX}=20\times \frac{3}{5}$
$V_{AX}=12\text{m/s}\dots \left(1\right)$
Now, velocity of $B$ along $x-axis$ along $AB$ is
$V_{BX}=V_E\cos {37}^{\circ }$
$V_{BX}=V_E\times \frac{4}{5}\dots \left(2\right)$
Now, from equation $\left(1\right)$ and $(2$)
$12=V_E\times \frac{4}{5}$
$V_E=\frac{60}{4}$
$V_E$ = $15$
Hence, the velocity $V_E$ is $15\text{m/s}$